Question

A charged capacitor is connected to an ideal inductor to form an LC circuit with a frequency of oscillation f = 1.6 Hz. At time t = 0 the capacitor is fully charged. At a given instant later the charge on the capacitor is measured to be 3.0 μC and the current in the circuit is equal to 75 μA. What is the maximum charge of the capacitor?

Answer 1

`8.0\mu C`

Step-by-step explanation:

We are given that

`f=1.6 Hz`

`q=3.0\mu C=3.0* 10^(-6) C`

`1\mu C=10^(-6) C`

Current,I=
`75\mu A=75* 10^(-6) A`

`1\mu A=10^(-6) A`

We have to find the maximum charge of the capacitor.

Charge on the capacitor,
`q=q_0cos\omega t`

`\omega=2\pi f=2\pi* 1.6=3.2\pi rad/s`

`3* 10^(-6)=q_0cos3.2\pi t`....(1)

`I=(dq)/(dt)=-q_0\omega sin\omega t`

`75* 10^(-6)=-q_0(3.2\pi)sin3.2\pi t`....(2)

Equation (2) divided by equation (1)

`-3.2\pi tan3.2\pi t=(75* 10^(-6))/(3* 10^(-6))=25`

`tan3.2\pi t=-(25)/(3.2\pi)=-2.488`

`3.2\pi t=tan^(-1)(-2.488)=-1.188rad`

`q_0=(q)/(cos\omega t)=(3* 10^(-6))/(cos(-1.188))=8.0* 10^(-6)=8\mu C`

Hence, the maximum charge of the capacitor=
`8.0\mu C`